Ta có :
\(3A=1+\frac{1}{3}+.....+\frac{1}{3^{98}}\)
\(\Rightarrow3A-A=\left(1+\frac{1}{3}+....+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{99}}\right)\)
\(\Rightarrow2A=1-\frac{1}{99}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{198}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)
=> 3A = 1 + 1/3 + 1/32 +... +1/398
=> 2A = 1 - 1/399
=> A = \(\frac{1-\frac{1}{3^{99}}}{2}\)
Mà \(1-\frac{1}{3^{99}}\) < 1 nên A < \(\frac{1}{2}\)
Ta có 3A=1/3^2+1/3^3+...+1/3^100
Mà A=1/3+1/3^2+...+1/3^99
suy ra 2A=-1/3+1/3^100
suy ra A=-1/3+1/3^100/2(ghi 1/3+1/3^100 rồi chia cho 2)
A=-1/6.vì -1/6<1/2
suy ra A<1/2(đpcm)
rất đầy đủ và chuẩn 99,99%
